Given that,

$$5\left( {{{\tan }^2}x - {{\cos }^2}x} \right) = 2\cos 2x + 9$$

$$ \Rightarrow 5\left( {{{{{\sin }^2}x} \over {{{\cos }^2}x}} - {{\cos }^2}x} \right) = 2\left( {2{{\cos }^2}x - 1} \right) + 9$$

Let $${\cos ^2}x = t,$$ then we have

$$5\left( {{{1 - t} \over t} - t} \right) = 2\left( {2t - 1} \right) + 9$$

$$ \Rightarrow 5\left( {{{1 - t - {t^2}} \over t}} \right) = 4t - 2 + 9$$

$$ \Rightarrow 5 - 5t - 5{t^2} = 4{t^2} + 7t$$

$$ \Rightarrow 9{t^2} + 12t - 5 = 0$$

$$ \Rightarrow 9{t^2} + 15t - 3t - 5 = 0$$

$$ \Rightarrow 3t\left( {3t + 5} \right) - 1\left( {3t + 5} \right) = 0$$

$$ \Rightarrow \left( {3t + 5} \right)\left( {3t - 1} \right) = 0$$

$$\therefore$$ $$t = {1 \over 3}$$ and $$t = - {5 \over 3}$$

If $$t = - {5 \over 3}$$ then $${\cos ^2}x$$ is negative

So , $$t$$ can not be $$ - {5 \over 3}$$.

So, correct value of $$t = {1 \over 3}$$ then $$\cos {}^2x = t = {1 \over 3}$$

$$\therefore\,\,\,$$ $$\cos 4x$$

$$ = 2{\cos ^2}2x - 1$$

$$ = 2{\left[ {2{{\cos }^2}x - 1} \right]^2} - 1$$

$$ = 2.{\left[ {2.{1 \over 3} - 1} \right]^2} - 1$$

$$ = 2.{\left( { - {1 \over 3}} \right)^2} - 1$$

$$ = {2 \over 9} - 1$$

$$ = - {7 \over 9}$$